The answer is 1. We'll use area consideration and Heron's formula.

Denote the sides a, b, c and the radius of the incircle as r. Denote the semiperimeter `s = 1 / 2 ( a + b + c ) . ` Then we see that the area of the triangle is `1 / 2 r a + 1 / 2 r b + 1 / 2 r c = r s ` while by Heron's formula it is `sqrt (s(s-a)(s-b)(s-c)) . ` Because of this, `r^2 = 1/s (s-a)(s-b)(s-c) .`

Let's also denote the length `AB_1 ` as `x, ` then `C B_1 = C A_1 = b - x , ` `B A_1 = B C_1 = a - b + x , ` `A C_1 = A B_1 = c - a + b - x = x ` (refer to the image).

This way `x = 1 / 2 ( c - a + b ) = s - a . ` The right triangle `A B_1 I` gives us `A I^2 = ( s - a )^2 + r^2 = ( s - a ) / s (s(s -a) + (s-b)(s-c)).`

The value inside the parentheses is equal to

`1/4 (a+b+c)(b+c-a) + 1/4 (a+c-b)(a+b-c) = 1/4 ((b+c)^2-a^2+a^2-(b-c)^2) = bc.`

Now we see that `AI^2 / bc = (s-a)/s, ` the same way `BI^2 / ac = (s-b)/s` and `CI^2 / ab = (s-c)/s. ` Thus, the sum in question is

`(s-a)/s+(s-b)/s+(s-c)/s = 1/s (3s-2s) = 1.`